Question

An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an observed specific rotation of 12.0°. If it is known that the specific rotation of the R enantiomer is –38.3°, determine the percentage of each isomer in the mixture.

Answer 1

Answer:

Percentage of R enantiomer = 34.33 %

Percentage of S enantiomer = 65.67 %

Explanation:

The formula for the calculation of the Optical rotation is:

$Specific\ rotation\ of\ mixture=(\frac {\%\ of\ R\ enantiomer}{100}\times {Specific\ rotation\ of\ R\ enantiomer})+(\frac {\%\ of\ S\ enantiomer}{100}\times {Specific\ rotation\ of\ S\ enantiomer})$

Given that:

For R enantiomer:

Let % = x %

Specific rotation = –38.3°

For S enantiomer:

% = 100  - x  

(2 enantiomers in mixture)

Specific rotation = 38.3° (Opposite direction of R enantiomer)

Given, Observed specific rotation of mixture = 12.0°

Thus,  

$12.0^0=\frac {x}{100}\times {(-38.3^0)}+\frac {100-x}{100}\times {(38.3^0)}$

Solving for x, we get that:

x = 34.33 %

Percentage of R enantiomer = 34.33 %

Percentage of S enantiomer = 65.67 %